如图C、D、E将线段AB分为四部分,用AC:CD:DE:EB=2:3:4:5,M、P、Q、N分别是AC、CD、DE、EB
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AC:CD:DE:EB=2:3:4:5

设AC=2b,则CD=3b,DE=4b,EB=5b,AB=14b

M为AC中点,则AM=AC/2=b

P为CD中点,则PD=CD/2=3b/2

Q为DE中点,则DQ=DE/2=2b

N为BE中点,则BN=EB/2=5b/2

MN=AB-AM-BN=14b-b-5b/2=21

则b=2×21/21=2

∴PQ=PD+DQ=3b/2+2b=7b/2=7×2/2=7